Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having trouble with two steps in a calculation of

$$\int_0^\infty\left(\frac{\sin x}{x}\right)^2\ dx$$

in a book.

  • They take the contours $C_R$ composed of upper half-circles $H_r=\{re^{i\theta}\,:\,\theta\in[0,\pi]\}$ and $G_R=\{Re^{i\theta}\,:\,\theta\in[0,\pi]\},$ where $r<R,$ and the real intervals $I_{r,R}=[-R,-r]$ and $J_{r,R}=[r,R].$

  • They notice that $2\sin^2x=\Re(1-e^{2ix})$ for $x\in\Bbb R.$ This allows them to consider the function $$f(z)=\frac{1-e^{2iz}}{z^2}.$$

  • Now they do something I don't understand. They say that $$\int_{I_{r,R}}f(z)\ dz=\int_{J_{r,R}}f(z)\ dz.$$ How is this true? I have $$\int_{I_{r,R}}f(z)\ dz=\int_{-R}^{-r}\frac{1-e^{2ix}}{x^2}\ dx=-\int_{R}^{r}\frac{1-e^{-2ix}}{x^2}\ dx=\int_{r}^{R}\frac{1-e^{-2ix}}{x^2}\ dx,$$ but $$\int_{J_{r,R}}f(z)\ dz=\int_{r}^{R}\frac{1-e^{2ix}}{x^2}\ dx$$ without the minus in the exponent... And the exponential function isn't even.

  • Now they say that $$|\int_{G_R}f(z)|\leq\frac{2\pi}R\to 0$$ as $R\to\infty.$ I've been looking at it for over an hour, and I can't see this. Maybe I won't post my calculations because the post is already long. It looks completely false to me. Shouldn't the exponential function in the numerator dominate the denominator? I tried to use the triangle inequality and the inequality $|e^z|\leq e^{|z|}$ after substituting $z=Re^{i\theta}.$

  • Now it's easy to finish the calculation, but since I don't understand the previous steps, it doesn't count.

share|cite|improve this question
up vote 3 down vote accepted

The two integrals (over $I_{r,R}$ and $J_{r,R}$) don't look equal to me, but their real parts will be equal, since the real part of $e^{2iz}$ is even. That is probably all you need (I haven't checked in detail, but remember you are starting with real integral).

In the upper half plane $e^{2iz}=e^{2ix}\cdot e^{-2y}$. The first factor has modulus $1$, the second is less than $1$. I think that deals with your second problem.

share|cite|improve this answer
Thank you. I can finish this now. It must be a mistake in the book then. They explicitly use the whole function and not its real part. – Bartek Feb 5 '13 at 7:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.